Rectangular Footing Sample Problem 264id
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RECTANGULAR FOOTING Problem
A 450 mm square interior column carries a dead load of 825 kN and a live load of 668 kN. A rectangular footing is required to carry the column loads such that the length of the long side must be equal to twice the width of the short side. Assume base of footing is 1.5 m below the ground surface. Allowable bearing pressure of soil is 192 kPa. fc’=27.6 MPa, fy=415 MPa, weight of 3 3 soil is 15.74 kN/m and that of concrete is 23.5 kN/m .
DL = 825 kN/m LL = 668 kN/m
Determine the dimensions of the rectangular footing. Determine the longitudinal steel requirements. Determine the steel requirements in short direction. 0.90 m
1.5 m 0.60 m
Solution (
Dimensions of the rectangular footing: Assume total depth of 600 mm.
) 4.28
Effective soil pressure: (
)
(
) 2.14
Area required: 1.39 d=0.525
( )
[(
Ultimate upward soil pressure:
(
)
(
)(
Considering one way shear: ( )( )
(
)
( ) use 2.14 m x 4.28 m x 0.60 m rectangular footing
√
(
) ]
) √
(
(
√
)
(
√
)
4.28
)
) 450
Considering two-way shear (punching shear):
975
450
975
2.14
Longitudinal steel requirements:
(
)(
)(
)
4.28
( (
0.45 0.45
2.14
(
)( )
) )(
1.915
(
1.915
qu = 224.76 kPa
)
)
(
)
(
)(
using 25mm ϕ bars (
)
Use 10 – 25 mm ϕ bars
)
Reinforcement in short direction: ( (
4.28
)( )
)(
)
)(
)
0.845
(
(
)(
)(
)
) (
(
)
(
)
(
(
) (
)
n = 2.9 say 3 bars
Using 22 mm ϕ bars (
)
)
Rectangular Footings Design Procedure (through Sample Problem)
Design a rectangular spread footing to carry 235 kips dead load and 115 kips live load from an 18-in. square tied column that contains #9 bars. One dimension of the footing is limited to a maximum of 7 ft. The allowable soil pressure is 5500 psf. neglect the effect of overburden. Use fc’ = 3000 psi, fy = 40000 psi, and the strength method of the ACI Code.
(A) Determine plan of footing. Assume a footing depth of 2 ft, or 300 psf.
Space limitation prevents one dimension from exceeding 7 ft; thus
Try 7 ft x 9 ft 8 in. (Area = 67.7 sq ft).
(B) Determine depth required for shear strength. This footing may be long enough for one-way beam action to govern. In such a case, a direct solution for the effective depth d is reasonably practical. The factored column load is (
)
(
)
The net upward pressure under factored load condition is
Making the nominal shear strength Vn=Vc so that shear reinforcement is not required, which means ( √ ( )(
)
) ) ( )( )
( √ (
)
Total depth = 17.9 + 3 (cover) + 1 (estimated bar diameter) = 21.9 in. Try 22 in. for total depth. Check this depth for twoway shear action as a slab, using critical section with d = 18 in. [ (
)
( )]
For βc = 9.67/7 = 1.38 < 2, and b0/d = 4(18 + 18)/18 = 8 < 20 for a four-sided critical section,
√
[ (
√
)]( (
)
) [
]
A 450 mm square interior column carries a dead load of 825 kN and a live load of 668 kN. A rectangular footing is required to carry the column loads such that the length of the long side must be equal to twice the width of the short side. Assume base of footing is 1.5 m below the ground surface. Allowable bearing pressure of soil is 192 kPa. fc’=27.6 MPa, fy=415 MPa, weight of 3 3 soil is 15.74 kN/m and that of concrete is 23.5 kN/m .
DL = 825 kN/m LL = 668 kN/m
Determine the dimensions of the rectangular footing. Determine the longitudinal steel requirements. Determine the steel requirements in short direction. 0.90 m
1.5 m 0.60 m
Solution (
Dimensions of the rectangular footing: Assume total depth of 600 mm.
) 4.28
Effective soil pressure: (
)
(
) 2.14
Area required: 1.39 d=0.525
( )
[(
Ultimate upward soil pressure:
(
)
(
)(
Considering one way shear: ( )( )
(
)
( ) use 2.14 m x 4.28 m x 0.60 m rectangular footing
√
(
) ]
) √
(
(
√
)
(
√
)
4.28
)
) 450
Considering two-way shear (punching shear):
975
450
975
2.14
Longitudinal steel requirements:
(
)(
)(
)
4.28
( (
0.45 0.45
2.14
(
)( )
) )(
1.915
(
1.915
qu = 224.76 kPa
)
)
(
)
(
)(
using 25mm ϕ bars (
)
Use 10 – 25 mm ϕ bars
)
Reinforcement in short direction: ( (
4.28
)( )
)(
)
)(
)
0.845
(
(
)(
)(
)
) (
(
)
(
)
(
(
) (
)
n = 2.9 say 3 bars
Using 22 mm ϕ bars (
)
)
Rectangular Footings Design Procedure (through Sample Problem)
Design a rectangular spread footing to carry 235 kips dead load and 115 kips live load from an 18-in. square tied column that contains #9 bars. One dimension of the footing is limited to a maximum of 7 ft. The allowable soil pressure is 5500 psf. neglect the effect of overburden. Use fc’ = 3000 psi, fy = 40000 psi, and the strength method of the ACI Code.
(A) Determine plan of footing. Assume a footing depth of 2 ft, or 300 psf.
Space limitation prevents one dimension from exceeding 7 ft; thus
Try 7 ft x 9 ft 8 in. (Area = 67.7 sq ft).
(B) Determine depth required for shear strength. This footing may be long enough for one-way beam action to govern. In such a case, a direct solution for the effective depth d is reasonably practical. The factored column load is (
)
(
)
The net upward pressure under factored load condition is
Making the nominal shear strength Vn=Vc so that shear reinforcement is not required, which means ( √ ( )(
)
) ) ( )( )
( √ (
)
Total depth = 17.9 + 3 (cover) + 1 (estimated bar diameter) = 21.9 in. Try 22 in. for total depth. Check this depth for twoway shear action as a slab, using critical section with d = 18 in. [ (
)
( )]
For βc = 9.67/7 = 1.38 < 2, and b0/d = 4(18 + 18)/18 = 8 < 20 for a four-sided critical section,
√
[ (
√
)]( (
)
) [
]
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